how to make sure my string format must be like this :
locker_number=3,email=ucup#gmail.com,mobile_phone=091332771331,firstname=ucup
i want my string format `"key=value,"
how to make regex for check my string on ruby?
This regex will find what you're after.
\w+=.*?(,|$)
If you want to capture each pairing use
(\w+)=(.*?)(?:,|$)
http://rubular.com/r/A2ernIzQkq
The \w+ is one or more occurrences of a character a-z, 1-9, or an underscore. The .*? is everything until the first , or the end of the string ($). The pipe is or and the ?: tells the regex no to capture that part of the expression.
Per your comment it would be used in Ruby as such,
(/\w+=.*?(,|$)/ =~ my_string) == 0
You can use a regex like this:
\w+=.*?(,|$)
Working demo
You can use this code:
"<your string>" =~ /\w+=.*?(,|$)/
What about something like this? It's picky about the last element not ending with ,. But it doesn't enforce the need for no commas in the key or no equals in the value.
'locker_number=3,email=ucup#gmail.com,mobile_phone=091332771331,firstname=ucup' =~ /^([^=]+=[^,]+,)*([^=]+=[^,]+)$/
Related
I am building a Rails 5.2 app.
In this app I got outputs from different suppliers (I am building a webshop).
The name of the shipping provider is in this format:
dhl_freight__233433
It could also be in this format:
postal__US-320202
How can I remove all that is before (and including) the __ so all that remains are the things after the ___ like for example 233433.
Perhaps some sort of RegEx.
A very simple approach would be to use String#split and then pick the second part that is the last part in this example:
"dhl_freight__233433".split('__').last
#=> "233433"
"postal__US-320202".split('__').last
#=> "US-320202"
You can use a very simple Regexp and a ask the resulting MatchData for the post_match part:
p "dhl_freight__233433".match(/__/).post_match
# another (magic) way to acces the post_match part:
p $'
Postscript: Learnt something from this question myself: you don't even have to use a RegExp for this to work. Just "asddfg__qwer".match("__").post_match does the trick (it does the conversion to regexp for you)
r = /[^_]+\z/
"dhl_freight__233433"[r] #=> "233433"
"postal__US-320202"[r] #=> "US-320202"
The regular expression matches one or more characters other than an underscore, followed by the end of the string (\z). The ^ at the beginning of the character class reads, "other than any of the characters that follow".
See String#[].
This assumes that the last underscore is preceded by an underscore. If the last underscore is not preceded by an underscore, in which case there should be no match, add a positive lookbehind:
r = /(?<=__[^_]+\z/
This requires the match to be preceded by two underscores.
There are many ruby ways to extract numbers from string. I hope you're trying to fetch numbers out of a string. Here are some of the ways to do so.
Ref- http://www.ruby-forum.com/topic/125709
line.delete("^0-9")
line.scan(/\d/).join('')
line.tr("^0-9", '')
In the above delete is the fastest to trim numbers out of strings.
All of above extracts numbers from string and joins them. If a string is like this "String-with-67829___numbers-09764" outut would be like this "6782909764"
In case if you want the numbers split like this ["67829", "09764"]
line.split(/[^\d]/).reject { |c| c.empty? }
Hope these answers help you! Happy coding :-)
I have a string from which I want to extract a certain part:
Original String: /abc/d7_t/g-12/jkl/m-n3/pqr/stu/vwx
Result Desired: /abc/d7_t/g-12/jkl/
The number of characters can vary in the entire string. It has alphabets, numbers, underscore and hyphen. I want to basically cut the string after the 5th "/"
I tried a few regex, but it seems there is some mistake with the format.
If a non-regexp approach is acceptable, how about this:
s.split('/').take(n).join('/')+'/'
Where s if your string (in your case: /abc/d7_t/g-12/jkl/m-n3/pqr/stu/vwx).
def cut_after(s, n)
s.split('/').take(n).join('/')+'/'
end
Then
cut_after("/abc/d7_t/g-12/jkl/m-n3/pqr/stu/vwx", 5)
should work. Not as compact as a regexp, but some people may find it clearer.
The regexp would be: %r(/(?:[^/]+/){4}). Note that it is a good idea in this case to use the %r literal version to avoid escaping slashes. Unescaped slashes are likely the cause of your format errors.
Match any sequence of chars except '/' 4 times :-
(\/[^\/]+){4}\/
I would like to use regular expression to check if my string have the format like following:
mc_834faisd88979asdfas8897asff8790ds_oa_ids
mc_834fappsd58979asdfas8897asdf879ds_oa_ids
mc_834faispd8fs9asaas4897asdsaf879ds_oa_ids
mc_834faisd8dfa979asdfaspo97asf879ds_dv_ids
mc_834faisd111979asdfas88mp7asf879ds_dv_ids
mc_834fais00979asdfas8897asf87ggg9ds_dv_ids
The format is like mc_<random string>_oa_ids or mc_<random string>_dv_ids . How can I check if my string is in either of these two formats? And please explain the regular expression. thank you.
That's a string start with mc_, while end with _oa_ids or dv_ids, and have some random string in the middle.
P.S. the random string consists of alpha-beta letters and numbers.
What I tried(I have no clue how to check the random string):
/^mc_834faisd88979asdfas8897asff8790ds$_os_ids/
Try this.
^mc_[0-9a-z]+_(dv|oa)_ids$
^ matches at the start of the line the regex pattern is applied to.
[0-9a-z] matces alphabetic and numeric chars.
+ means that there should be one or more chars in this set
(dv|oa) matches dv or oa
$ matches at the end of the string the regex pattern is applied to.
also matches before the very last line break if the string ends with a line break.
Give /\Amc_\w*_(oa|dv)_ids\z/ a try. \A is the beginning of the string, \z the end. \w* are one or more of letters, numbers and underscores and (oa|dv) is either oa or dv.
A nice and simple way to test Ruby Regexps is Rubular, might have a look at it.
This should work
/mc_834([a-z,0-9]*)_(oa|dv)_ids/g
Example: http://regexr.com?2v9q7
Suppose I have string variables like following:
s1="10$"
s2="10$ I am a student"
s3="10$Good"
s4="10$ Nice weekend!"
As you see above, s2 and s4 have white space(s) after 10$ .
Generally, I would like to have a way to check if a string start with 10$ and have white-space(s) after 10$ . For example, The rule should find s2 and s4 in my above case. how to define such rule to check if a string start with '10$' and have white space(s) after?
What I mean is something like s2.RULE? should return true or false to tell if it is the matched string.
---------- update -------------------
please also tell the solution if 10# is used instead of 10$
You can do this using Regular Expressions (Ruby has Perl-style regular expressions, to be exact).
# For ease of demonstration, I've moved your strings into an array
strings = [
"10$",
"10$ I am a student",
"10$Good",
"10$ Nice weekend!"
]
p strings.find_all { |s| s =~ /\A10\$[ \t]+/ }
The regular expression breaks down like this:
The / at the beginning and the end tell Ruby that everything in between is part of the regular expression
\A matches the beginning of a string
The 10 is matched verbatim
\$ means to match a $ verbatim. We need to escape it since $ has a special meaning in regular expressions.
[ \t]+ means "match at least one blank and/or tab"
So this regular expressions says "Match every string that starts with 10$ followed by at least one blank or tab character". Using the =~ you can test strings in Ruby against this expression. =~ will return a non-nil value, which evaluates to true if used in a conditional like if.
Edit: Updated white space matching as per Asmageddon's suggestion.
this works:
"10$ " =~ /^10\$ +/
and returns either nil when false or 0 when true. Thanks to Ruby's rule, you can use it directly.
Use a regular expression like this one:
/10\$\s+/
EDIT
If you use =~ for matching, note that
The =~ operator returns the character position in the string of the
start of the match
So it might return 0 to denote a match. Only a return of nil means no match.
See for example http://www.regular-expressions.info/ruby.html on a regular expression tutorial for ruby.
If you want to proceed to cases with $ and # then try this regular expression:
/^10[\$#] +/
I'm trying to remove any lines that begin with the character '>' in a long string (i.e. replies to an email).
In PHP I'd iterate over each line with an if statement, in linux I'd try and use sed or awk.
What's the most elegant rails approach?
You can try this:
your_string.gsub(/^\>.+\n/,'')
Your question is implying that the input is one string, containing multiple lines.
Do you want the output to be just one string with multiple lines as well? I'm assuming yes.
either using String and Array operations:
str.lines.reject{|x| x =~ /^>/}.join # this will return a new string, without those ">" lines
or using Regular Expressions:
str.gsub(/^>.+\n*/. '')
Better Solution:
You will need to use non-greedy multi-line matching mode for your Regular Expression:
str.gsub(/^>.*?$\n*/m, '') # by using gsub!() you can modify the string in place
^> matches your ">" character at the start of a line
.*?$ matches any characters after the start character until the end of the line (non-greedy)
\n* matches the newline character itself if any (you want to remove that as well)
the "m" at the end of the regular expressions indicates multi-line matching , which will apply the RegExp for each line in the string.
It should work as you expect:
your_string.lines.to_a.reject{|line| line[0] == '>'}.join